Question

The ${{E}_{a}}$ of reaction in presence of catalyst is $4.15KJmo{{l}^{-1}}$ and in absence of catalyst is $8.3KJmo{{l}^{-1}}$. What is the slope of the plot $lnK$ vs $\dfrac{1}{T}$ in the absence of a catalyst?
A. +1
B. -1
C. +1000
D. -1000

Answer 1

The minimum amount of extra energy required for a reacting molecule to get converted into product is known as activation energy of that molecule. It can also be described as the minimum amount of energy needed to activate or energize molecules or atoms so that they can undergo a chemical reaction or transformation.

Complete step by step solution:
Activation energy is generally represented by ${{E}_{a}}$ it is generally depend on two main factors which are nature of the reactants; in case of ionic reactant the value of activation energy will be low because there is an attraction between reacting species. While in the case of covalent reactant the value of ${{E}_{a}}$ will be high because energy is required to break the older bonds. Another factor is effect of catalyst; ${{E}_{a}}$ value will be low in case of positive catalyst as it provides an alternate path while the negative catalyst provides such path in which the value of ${{E}_{a}}$ will be high.
We know the equation of activation energy is given by:
$\ln K=\ln A-\dfrac{{{E}_{a}}}{R}\times \dfrac{1}{T}$
Where slope is shown by $-\dfrac{{{E}_{a}}}{R}$like the equation y = mx + c where m is slope
Hence ${{E}_{a}}$in absence of catalyst = 8.3 KJ = $8.3\times 1000J$
While the value of R constant = 8.314
$\therefore \dfrac{-Ea}{R}=\dfrac{-8.3\times 1000}{8.3}=-1000$

Hence option D is the correct answer.

Note: A catalyst is a chemical substance that either increases or decreases the rate of a chemical reaction. In the case of activation energy a catalyst lowers the value. However the energies of the original reactants remain the same. A catalyst only alters the activation energy.